MD5 - With Example

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MD5 Algorithm Descrip on with an example

We begin by supposing that we have a b‐bit message as input, and that we wish to find its message digest. Here b is an arbitrary nonnegative integer; b may be zero, it need not be a multiple of eight, and it may be arbitrarily large. We imagine the bits of the message written down as follows: m_0 m_1 ... m_{b‐1} The following five steps are performed to compute the message digest of the message. Step 1. Append Padding Bits

Ex. : The message should be "Hello World!", which is of length 12 bytes, they are…

The message is "padded" (extended) so that its length (in bits) is congruent to 448, modulo 512. That is, the message is extended so that it is just 64 bits shy of being a mul ple of 512 bits long. Padding is always performed, even if the length of the message is already congruent to 448, modulo 512. Padding is performed as follows: a single "1" bit is appended to the message, and then "0" bits are appended so that the length in bits of the padded message becomes congruent to 448, modulo 512. In all, at least one bit and at most 512 bits are appended.

Ex. In this case, the message fits a single 512 bit block, so we will add 448 – 12*8 = 352 padding bits (it will start with “1” then 351 “0”s)

Step 2. Append Length

A 64‐bit representation of b (the length of the message before the padding bits were added) is

appended to the result of the previous step. In the unlikely event that b is greater than 2^64, then only the low‐order 64 bits of b are used. (These bits are appended as eight bytes and appended low‐order byte)

Ex. : The length of message in bits is 96 bits == 00000000 00000000 00000000 00000000 00000000 00000000 00000000 01100000… Taking low order byte first will result in

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At this point the resulting message (after padding with bits and with b) has a length that is an exact mul ple of 512 bits. Equivalently, this message has a length that is an exact mul ple of 16 (32‐bit) words. Let M[0 ... N‐1] denote the words of the resul ng message, where N is a mul ple of 16. Ex. Now the only block available holds…

Step 3. Ini alize MD Buﬀer A four‐word buﬀer (A,B,C,D) is used to compute the message digest. Here each of A, B, C, D is a 32‐bit register. These registers are initialized to the following values in hexadecimal, low‐order bytes first): 32‐bit registers values (Hexa) After conversion to in low‐order bytes first (Hexa) After Conversion to in low‐order bytes first (Decimal)

word A: 01 23 45 67

word B: 89 ab cd ef

word C: fe dc ba 98

word D: 76 54 32 10

0x67452301

0xefcdab89

0x98badcfe

0x10325476

a: 1732584193

b: 4023233417

c: 2562383102

d: 271733878

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Step 4. Process Message in 16‐Word Blocks We first define four auxiliary func ons that each take as input three 32‐bit words and produce as output one 32‐bit word. F(X,Y,Z) = XY v not(X) Z G(X,Y,Z) = XZ v Y not(Z) H(X,Y,Z) = X xor Y xor Z I(X,Y,Z) = Y xor (X v not(Z)) In each bit position F acts as a conditional: if X then Y else Z. The function F could have been defined using + instead of v since XY and not(X)Z will never have 1's in the same bit position.) It is interesting to note that if the bits of X, Y, and Z are independent and unbiased, the each bit of F(X,Y,Z) will be independent and unbiased. The functions G, H, and I are similar to the function F, in that they act in "bitwise parallel" to produce their output from the bits of X, Y, and Z, in such a manner that if the corresponding bits of X, Y, and Z are independent and unbiased, then each bit of G(X,Y,Z), H(X,Y,Z), and I(X,Y,Z) will be independent and unbiased. Note that the function H is the bit‐wise "xor" or "parity" function of its inputs. This step uses a 64‐element table T[1 ... 64] constructed from the sine func on. Let T[i] denote the i‐th element of the table, which is equal to the integer part of 4294967296 mes abs(sin(i)), where i is in

radians

(switch your calculator to Radians mode instead of Degrees mode). The elements of the table are given in the appendix. Do the following:

/* Process each 16-word block. */ For i = 0 to N/16-1 do

/* Copy block i into X. */

Ex. Now “X” the 16 words each of 32 bits will hold, as the following loop does

For j = 0 to 15 do

Set X[j] to M[i*16+j]. end /* of loop on j */

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/* Save A as AA, B as BB, C as CC, and D as DD. */ AA = A BB = B CC = C DD = D /* Round 1. */

/* Let [abcd k s i] denote the operation

a = b + ((a + F(b,c,d) + X[k] + T[i]) <<< s). */ /* Do the following 16 operations. */

[ABCD 0 7 1]

Ex. * Perform in Binary: F(b,c,d) = F(B,C,D) = BC v not(B) D =

(4023233417^2562383102)V(NOT(4023233417)^ 271733878) = 2562383102

* (a + F(b,c,d) + X[k] + T[i]) = 1732584193+2562383102+1819043144+3614090360

=9728100799(10 01000011110101110000100110111111)

The above is 34 bits, so we do modular additions, to get the results of only 32 bits

9728100799 MOD (2Power32) = 1138166207 (01000011110101110000100110111111)

* ((a + F(b,c,d) + X[k] + T[i]) <<< s) = 1138166207 <<< 7 = 3951353761

OR 01000011110101110000100110111111 <<< 7 = 11101011100001001101111110100001

* Again Modular Addition with b: a = b + ((a + F(b,c,d) + X[k] + T[i]) <<< s) =

(4023233417+3951353761)MOD(2Power32) = 3679619882

* Now a is 3679619882. b, c, and d are not changed.

[DABC 1 12 2]

* Just as if you call [abcd k s i] by passing [DABC 1 12 2], meaning in the expression mentioned

in the comment above, you will replace a by D, b by A, c by B d by C, k by 1, s by 12, and i by 2

* Now c is 2562383102 and the rest of registers is unchanged

[CDAB 2 17 3] [BCDA 3 22 4]

[ABCD 4 7 5] [DABC 5 12 6] [CDAB 6 17 7] [BCDA 7 22 8] [ABCD 8 7 9] [DABC 9 12 10] [CDAB 10 17 11] [BCDA 11 22 12] [ABCD 12 7 13] [DABC 13 12 14] [CDAB 14 17 15] [BCDA 15 22 16]

/* Round 2. */

/* Let [abcd k s i] denote the operation

a = b + ((a + G(b,c,d) + X[k] + T[i]) <<< s). */ /* Do the following 16 operations. */

[ABCD 1 5 17] [DABC 6 9 18] [CDAB 11 14 19] [BCDA 0 20 20] [ABCD 5 5 21] [DABC 10 9 22] [CDAB 15 14 23] [BCDA 4 20 24] [ABCD 9 5 25] [DABC 14 9 26] [CDAB 3 14 27] [BCDA 8 20 28] [ABCD 13 5 29] [DABC 2 9 30] [CDAB 7 14 31] [BCDA 12 20 32]

/* Round 3. */

/* Let [abcd k s t] denote the operation

a = b + ((a + H(b,c,d) + X[k] + T[i]) <<< s). */ /* Do the following 16 operations. */

/* Round 4. */

/* Let [abcd k s t] denote the operation

a = b + ((a + I(b,c,d) + X[k] + T[i]) <<< s). */ /* Do the following 16 operations. */

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Round/Function Iteration

Changed

_Register

Value

1 / FF

1 a

3679619882

2 d

2081685101

3 c

133919401

4 b

2426329734

5 a

1168620452

6 d

4019849227

7 c

370807664

8 b

420445366

9 a

4198619016

10 d

2676785399

11 c

1993335089

12 b

1109401451

13 a

4048899853

14 d

3950087495

15 c

1946913403

16 b

2833700088

2 / GG

17 a

260718522

18 d

1355334187

19 c 80112311

20 b

3466307461

21 a

396033290

22 d

1763835213

23 c

4239641319

24 b

3052862749

25 a

2616217882

26 d

921013414

27 c

2418186433

28 b

2064486443

29 a

1988550793

30 d

146954281

31 c

198383918

32 b

1503938765

3 / HH

33 a

1829143082

34 d

2652512423

35 c

44713168

36 b

3452888323

37 a

4152347168

38 d

2590106416

39 c

4138040678

40 b

3946422880

41 a

1824650282

42 d

3516852032

43 c

823396667

44 b

206704551

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45 a

1129083802

46 d

1795018062

47 c

2728674280

48 b

2145583183

4 / II

49 a

3393064847

50 d

3765494986

51 c

2359501063

52 b

1432248893

53 a

449791293

54 d

411940788

55 c

548752943

56 b

1539193972

57 a

3430658958

58 d

3442781692

59 c

1141842014

60 b

924561567

61 a 538764524

62 d 2077978953

63 c 4184057696

64 b 1186497226

The last four rows hold the last values of the four registers…

/* Then perform the following additions. (That is increment each of the four registers by the value it had before this block was started.) */

A = A + AA B = B + BB C = C + CC D = D + DD

Ex. At last Modular Addition with the original values

A = A + AA = (1732584193 + 538764524 ) MOD(2Power32) = 2271348717 = 10000111 01100010

00000111 11101101 = 0x876207ED

B = B + BB = (4023233417 + 1186497226) MOD(2Power32) = 914763347 = 00110110 10000110

00101110 01010011 = 0x36862E53

C = C + CC = (2562383102 + 4184057696) MOD(2Power32) = 2451473502 = 10010010 00011110

10000100 01011110 = 0x921E845E

D = D + DD = (271733878 + 2077978953) MOD(2Power32) = 2349712831 = 10001100 00001101

11000101 10111111 = 0x8C0DC5BF

Ex. This is for the only first 512 bit block, now continue to the next 512-bit blocks. But this is not

in this example ☺. We are done manipulating our “Hello World!” message.

end /* of loop on i */

Step 5. Output

The message digest produced as output is A, B, C, D. That is, we begin with the low‐order byte of A, and end with the high‐order byte of D.

Ex. After we are done with the above loop, we take A, B, C, and D and reorder their bytes:

(Each register is 32 bits (4 Bytes))

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A 4 3 2 1

B 8 7 6 5

C 12 11 10 9

D 16 15 14 13

* Now the MD5 digest is stored as the above bytes in ascending order (1, 2, 3, …) (Remember

that a single byte is equivalent to 2 Hexadecimals)

* The MD5 digest for the above example is 0xED076287532E86365E841E92BFC50D8C

Summary The MD5 message‐digest algorithm is simple to implement, and provides a "fingerprint" or message digest of a message of arbitrary length. It is conjectured that the difficulty of coming up with two messages having the same message digest is on the order of 2^64 opera ons, and that the diﬃculty of coming up with any message having a given message digest is on the order of 2^128 opera ons. The MD5 algorithm has been carefully scru nized for weaknesses. It is, however, a relatively new algorithm and further security analysis is of course justified, as is the case with any new proposal of this sort. References